An alternating series identity with a hidden hyperbolic tangent How does one use the inverse Mellin transform to prove that the following identity holds?
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n(e^{n\pi} + 1)} = \frac{1}{8}(\pi - 5\log(2))$$
The identity follows from MO189199 and the penultimate identity on this page (for $x=\frac{1}{2}$).

Scratch-work: I computed the Mellin transform of 
$$f(x) = \frac{(-1)^x}{x(e^{x\pi} + 1)}$$
and re-wrote the function in terms of its inverse Mellin Transform as (substituting $x = n$)
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n(e^{n\pi} + 1)} = \frac{1}{2\pi i}\int_{C} (2 \pi)^{1-s} \Gamma(s-1)\big(\zeta(s-1, \frac{1}{2} - \frac{i}{2}) - \zeta(s-1, 1-\frac{i}{2})\big)\zeta(s)ds$$
But I am not altogether confident that this has been computed correctly, and I am still not sure how to find the poles and compute the residues. For example, checking for a residue at $s=0$ gives
$$\frac{1}{8}(\pi - 2\pi i)$$
There seems to be a nontrivial contribution at $s = 1$ and $s = -1$ as well, but already the mathematics has exceeded what I understand; I am especially unclear with regard to how one accounts for residues with respect to the generalized Riemann zeta function $\zeta(s,a)$.
Clarity on using this approach or another to demonstrate the main identity would be appreciated!
 A: Using$$\tanh\left(x\right)=1+2\underset{n=1}{\overset{\infty}{\sum}}\frac{\left(-1\right)^{n}}{e^{2nx}}$$
 we have$$\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n\left(e^{\pi n}+1\right)}=\frac{1}{2}\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}\left(1-\tanh\left(\frac{\pi n}{2}\right)\right)}{n}=\frac{1}{2}\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n}-\frac{1}{2}\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n}\tanh\left(\frac{\pi n}{2}\right)=$$
 $$=-\frac{1}{2}\log\left(2\right)-\frac{1}{2}\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n}\tanh\left(\frac{\pi n}{2}\right).$$
 Now we using the identity$$\tanh\left(\frac{\pi x}{2}\right)=\frac{4x}{\pi}\underset{k\geq1}{\sum}\frac{1}{\left(2k-1\right)^{2}+x^{2}}$$
 and so we have$$\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n\left(e^{\pi n}+1\right)}=-\frac{1}{2}\log\left(2\right)-\frac{2}{\pi}\underset{n\geq1}{\sum}\left(-1\right)^{n}\underset{k\geq1}{\sum}\frac{1}{\left(2k-1\right)^{2}+n^{2}}=-\frac{1}{2}\log\left(2\right)-\frac{2}{\pi}\underset{k\geq1}{\sum}\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{\left(2k-1\right)^{2}+n^{2}}.$$
 The last sum can be calculated$$\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{\left(2k-1\right)^{2}+n^{2}}=\frac{\left(2k-1\right)\pi-\sinh\left(\left(2k-1\right)\pi\right)}{2\left(2k-1\right)^{2}\sinh\left(\left(2k-1\right)\pi\right)}$$
 hence$$\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n\left(e^{\pi n}+1\right)}=-\frac{1}{2}\log\left(2\right)+\frac{1}{\pi}\underset{k\geq1}{\sum}\frac{1}{\left(2k-1\right)^{2}}-\underset{k\geq1}{\sum}\frac{1}{\left(2k-1\right)\sinh\left(\left(2k-1\right)\pi\right)}.$$
 Obviously$$\underset{k\geq1}{\sum}\frac{1}{\left(2k-1\right)^{2}}=\frac{1}{8}\pi^{2}$$
 and using$$\frac{\pi}{\sinh\left(\pi x\right)}=\frac{2\pi}{e^{\pi x}-e^{-\pi x}}=x\underset{m\in\mathbb{Z}}{\sum}\frac{\left(-1\right)^{m}}{x^{2}+m^{2}}$$
 we have$$\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n\left(e^{\pi n}+1\right)}=-\frac{1}{2}\log\left(2\right)+\frac{1}{8}\pi-\frac{1}{\pi}\underset{k\geq1}{\sum}\underset{m\in\mathbb{Z}}{\sum}\frac{\left(-1\right)^{m}}{\left(2k-1\right)^{2}+m^{2}}$$
 and the last sum is (you can find the proof here closed form for a double sum) is$$-\frac{1}{\pi}\underset{k\geq1}{\sum}\underset{m\in\mathbb{Z}}{\sum}\frac{\left(-1\right)^{m}}{\left(2k-1\right)^{2}+m^{2}}=-\frac{\log\left(2\right)}{8}$$
 and this complete the proof.
A: The following approach does not use the Mellin transform, but it is worth mentioning. 
It is quite easy to prove that:
$$\sum_{n=1}^{+\infty}\frac{\cos(\pi n)}{n} e^{-nk\pi} = -\log\left(1+e^{-k\pi}\right)$$
hence the original sum equals:
$$-\log\prod_{k=1}^{+\infty}\frac{1+e^{-(2k-1)\pi}}{1+e^{-2k\pi}}.$$
The last product is clearly related with the Jacobi triple product.
In particular, since the values of the Jacobi theta functions $\vartheta_3$ and $\vartheta_4$ in $q=e^{-\pi}$ are known, the claim simply follows.
